Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am not able to prove this for sure by myself... To be more precise, $A$ is a $n \times n$ matrix of rank $n-1$ such that all diagonal elements of A are positive, off-diagonal elements can be positive or negative, and its last row is the weighted sum of all the other rows ($i.e. A(n,:) = -1/n \sum_{i=1}^{n-1} i*A(i,:)$), which explains why $rk(A) = n-1$. I would then like to know if is possible to say that the matrix $\alpha I + \beta A $ with $\alpha >0$ and $\beta >0$ is invertible ? It seems to me that this is the case in general, though I could not proof that it was true for any $\alpha >0$ and $\beta >0$, any idea ?

share|cite|improve this question
up vote 0 down vote accepted

The matrix $\alpha I+\beta A$ is invertible iff $-\alpha/\beta$ is not an eigenvalue of $A$.

share|cite|improve this answer
thank you this is what I was looking for – Tibo Mar 5 '13 at 23:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.